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Komok [63]
3 years ago
9

The two forces acting on a skydiver are __________ and air _____________.

Physics
2 answers:
Ivan3 years ago
8 0

Answer:

Gravity and

Air resistance

Explanation:

Stolb23 [73]3 years ago
3 0

Answer:

Gravity and

Air resistance

Explanation:

The two forces acting on a skydiver are gravitational force and air resistance.

Gravitational force is a force that tends to pull all massive bodies towards the center of the earth. It works on all bodies that has mass. The larger or bigger the mass, the more the pull of gravity on the body.

Air resistance is the drag of air on a body as it passes to it. It is resisting force.

  • When a sky diver jumps out of a plane, he/she encounters both gravity and air resistance.
  • It soon balances both force and attain terminal velocity.
  • Air resistance is a frictional force that opposes motion.
  • This frictional force pushes in the opposite direction of motion
  • Motion direction is downward due to the celerity caused by gravity.
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The half-life of an anticancer drug is 170 days at 25 oC. Degradation follows the first-order kinetics. When the drug is stored
julsineya [31]

Answer:

1694 days

Explanation:

In first-order kinetics, the rate is proportional to the amount.

dA/dt = kA

For first-order kinetics, the rate k can be found using the half-life:

t₁,₂ = (ln 2) / k

In other words, the half-life is inversely proportional with the rate.

At the lower temperature, the rate is reduced to a third, so the half-life increases by a factor of 3.  Meaning that the new half-life is 170 × 3 = 510 days.

The "shelf life" is the time it takes to reduce the initial amount to 10%.  We can solve for this using the half-life equation.

A = A₀ (½)^(t / t₁,₂)

A₀/10 = A₀ (½)^(t / 510)

1/10 = (½)^(t / 510)

ln(1/10) = (t / 510) ln(½)

ln(10) = (t / 510) ln(2)

ln(10) / ln(2) = t / 510

t = 510 ln(10) / ln(2)

t ≈ 1694

5 0
3 years ago
2. A jack exerts a vertical force of 4.5 X 103
skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

\\  \\

Given :-

\star  \sf  \small force = 4.5 \times  {10}^{3}  \: newton

\star  \sf  \small distance = 0.25 \: meter

\\  \\

To find:-

\sf \star \: work = \: ?

\\  \\

Solution:-

we know :-

\bf \dag \boxed{ \rm work = force \times distance}

\\  \\

So:-

\dashrightarrow \sf work = force \times distance

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{0 \cancel.5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times  \cancel \frac{5}{10}  \\

\\  \\

\dashrightarrow \sf work =  \dfrac{4\cancel.5}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10}  \times 1 {0}^{3} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{3 - 1} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{10 {}^{0} }  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45}{1}  \times 1 {0}^{2} \times  \dfrac{1}{2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times  \cancel{10}}{ \cancel2} \\

\\  \\

\dashrightarrow \sf work =  \dfrac{45 \times 10 \times 5}{ 1} \\

\\  \\

\dashrightarrow \sf work =225 \times 10

\\  \\

\dashrightarrow \bf work =\red{2250\: joule}

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